complex infinite series

[logistic_guy]

here is the question

Use multiplication of series to show that \(\displaystyle \frac{e^z}{z(z^2 + 1)} = \frac{1}{z} + 1 - \frac{1}{2}z - \frac{5}{6}z^2 + \cdots \ (0 < |z| < 1)\).


my attemb
i think it want me to find infinite series for \(\displaystyle e^z\) and \(\displaystyle \frac{1}{z}\) and \(\displaystyle \frac{1}{z^2 + 1}\) then multiply them together
i know how to find taylor series for \(\displaystyle e^z\)
but i don't know how to find series for \(\displaystyle \frac{1}{z}\) and \(\displaystyle \frac{1}{z^2 + 1}\)

 

[fresh_42]

You can leave [imath] 1/z [/imath] as it is. For [imath] 1/(1+z^2) [/imath] use long division or look it up on Wolfram Alpha, or build the Taylor series by differentiations.

 

[logistic_guy]

You can leave [imath] 1/z [/imath] as it is. For [imath] 1/(1+z^2) [/imath] use long division or look it up on Wolfram Alpha, or build the Taylor series by differentiations.

i don't understand what you mean by long division
how to do division when i've only \(\displaystyle 1\) in the numator?
i thought long division is done when the numator have higher degree

 

[topsquark]

i don't understand what you mean by long division
how to do division when i've only \(\displaystyle 1\) in the numator?
i thought long division is done when the numator have higher degree

Google polynomial division.

-Dan

 

[fresh_42]

i don't understand what you mean by long division
how to do division when i've only \(\displaystyle 1\) in the numator?
i thought long division is done when the numator have higher degree

I don't care. Nike: Just do it!

[math]\begin{array}{lll} 1:(1+z^2)=1 - z^2 + z^4\mp \text{ etc.}\\[6pt] -(1+z^2)\\[6pt] ----------\\[6pt] -z^2\\[6pt] -(-z^2-z^4)\\[6pt] -------------\\[6pt] z^4\\[6pt] -(z^4+z^6)\\[6pt] -------------\\[6pt] -z^6\\[6pt] \text{etc.} \end{array}[/math]
The only thing that cannot be divided by is zero.

 

[logistic_guy]

Google polynomial division.

-Dan

i know how to do it but not with \(\displaystyle 1\) in the numator

I don't care. Nike: Just do it!

[math]\begin{array}{lll} 1:(1+z^2)=1 - z^2 + z^4\mp \text{ etc.}\\[6pt] -(1+z^2)\\[6pt] ----------\\[6pt] -z^2\\[6pt] -(-z^2-z^4)\\[6pt] -------------\\[6pt] z^4\\[6pt] -(z^4+z^6)\\[6pt] -------------\\[6pt] -z^6\\[6pt] \text{etc.} \end{array}[/math]
The only thing that cannot be divided by is zero.

why the first step in the long division, you divide \(\displaystyle 1\) by \(\displaystyle 1\) and not \(\displaystyle 1\) by \(\displaystyle z^2\)?

 

[topsquark]

i know how to do it but not with \(\displaystyle 1\) in the numator


why the first step in the long division, you divide \(\displaystyle 1\) by \(\displaystyle 1\) and not \(\displaystyle 1\) by \(\displaystyle z^2\)?

Honestly. You've never seen this before? Have you taken a course in complex Calculus?

-Dan

 

[fresh_42]

i know how to do it but not with \(\displaystyle 1\) in the numator


why the first step in the long division, you divide \(\displaystyle 1\) by \(\displaystyle 1\) and not \(\displaystyle 1\) by \(\displaystyle z^2\)?

Yes, because diving by [imath] z^2 [/imath] leads nowhere. The division by [imath] 1 [/imath] keeps it going. I can divide by whatever I want to, except zero, of course.

You can also formalize it by the Euclidean algorithm:
[math]\begin{array}{lll} 1&=1\cdot (1+z^2) - z^2\\[6pt] -z^2&=-z^2\cdot (1+z^2) +z^4\\[6pt] z^4&=z^4\cdot (1+z^2)-z^6\\[6pt] -z^6&=-z^6\cdot (1+z^2) +z^8\\[6pt] \text{etc.} \end{array}[/math]but the long division is easier and more known. The algorithm of dividing a polynomial [imath] p(z) [/imath] by a polynomial [imath] q(z) [/imath]
is always [imath] p(z)=s(z)\cdot q(z)+r(z) [/imath] no matter what [imath] p(z),q(z) [/imath] are. The example is a bit of a downside-up procedure in this specific case since the degree of [imath] r(z) [/imath] becomes larger rather than smaller. This means mathematically that the series is only a valid approximation for small values of [imath] z [/imath] around [imath] z=0. [/imath]

If you look at the thread you might observe that I already changed from your [imath] z^2+1 [/imath] in post #1 to my [imath] 1+z^2 [/imath] in post #2. That was on purpose!

In case you insist of the usual measure that the degree of the remainder [imath] r(z) [/imath] in Euclid's algorithm becomes smaller and smaller, then you have to transform the variable [imath] z \mapsto 1/x [/imath] and divide
[math] \dfrac{1}{1+z^2}=\dfrac{1}{1+\dfrac{1}{x^2}}=\dfrac{1}{\dfrac{x^2+1}{x^2}}=\dfrac{x^2}{x^2+1} [/math]instead and re-transform it by [imath] x\mapsto 1/z [/imath] afterward. I simply avoided these unnecessary steps by dividing [imath] 1:(1+z^2). [/imath]

 

[logistic_guy]

Honestly. You've never seen this before? Have you taken a course in complex Calculus?

-Dan

i seen it but i don't think to calculate it

Yes, because diving by [imath] z^2 [/imath] leads nowhere. The division by [imath] 1 [/imath] keeps it going. I can divide by whatever I want to, except zero, of course.

You can also formalize it by the Euclidean algorithm:
[math]\begin{array}{lll} 1&=1\cdot (1+z^2) - z^2\\[6pt] -z^2&=-z^2\cdot (1+z^2) +z^4\\[6pt] z^4&=z^4\cdot (1+z^2)-z^6\\[6pt] -z^6&=-z^6\cdot (1+z^2) +z^8\\[6pt] \text{etc.} \end{array}[/math]but the long division is easier and more known. The algorithm of dividing a polynomial [imath] p(z) [/imath] by a polynomial [imath] q(z) [/imath]
is always [imath] p(z)=s(z)\cdot q(z)+r(z) [/imath] no matter what [imath] p(z),q(z) [/imath] are. The example is a bit of a downside-up procedure in this specific case since the degree of [imath] r(z) [/imath] becomes larger rather than smaller. This means mathematically that the series is only a valid approximation for small values of [imath] z [/imath] around [imath] z=0. [/imath]

If you look at the thread you might observe that I already changed from your [imath] z^2+1 [/imath] in post #1 to my [imath] 1+z^2 [/imath] in post #2. That was on purpose!

In case you insist of the usual measure that the degree of the remainder [imath] r(z) [/imath] in Euclid's algorithm becomes smaller and smaller, then you have to transform the variable [imath] z \mapsto 1/x [/imath] and divide
[math] \dfrac{1}{1+z^2}=\dfrac{1}{1+\dfrac{1}{x^2}}=\dfrac{1}{\dfrac{x^2+1}{x^2}}=\dfrac{x^2}{x^2+1} [/math]instead and re-transform it by [imath] x\mapsto 1/z [/imath] afterward. I simply avoided these unnecessary steps by dividing [imath] 1:(1+z^2). [/imath]

thank for this information. i've a point and i want to emphasize on it

you say

I can divide by whatever I want to, except zero, of course.

if i divide by \(\displaystyle z^2\) i get

\(\displaystyle \frac{1}{z^2 + 1} = \frac{1}{z^2} - \frac{1}{z^4} + \frac{1}{z^6} - \cdots\)

if i divide by \(\displaystyle 1\) i get

\(\displaystyle \frac{1}{z^2 + 1} = 1 - z^2 + z^4 - \cdots\)

why there's two different series and how to know i must chose the second series?

 

[fresh_42]

Your series approximates the quotient at [imath] z=\infty [/imath] i.e. for very large values, mine approximates the quotient around [imath] z=0. [/imath] Since you want to multiply three series, you should consider the same ranges. You can also use
[math] 1/z = \sum_{n=0}^\infty (-1)^n (-1 + z)^n [/math]for [imath] |-1+z|<1 [/imath] if you want to avoid any powers of [imath] z [/imath] with negative exponents which I assumed.

 

[topsquark]

i seen it but i don't think to calculate it

You think going back and looking at complex series is embarrassing, too?

-Dan

 

[logistic_guy]

You think going back and looking at complex series is embarrassing, too?

-Dan

not really

Your series approximates the quotient at [imath] z=\infty [/imath] i.e. for very large values, mine approximates the quotient around [imath] z=0. [/imath] Since you want to multiply three series, you should consider the same ranges. You can also use
[math] 1/z = \sum_{n=0}^\infty (-1)^n (-1 + z)^n [/math]for [imath] |-1+z|<1 [/imath] if you want to avoid any powers of [imath] z [/imath] with negative exponents which I assumed.

how many terms do i take from each infinie series
\(\displaystyle \frac{1}{z} = 1 + 1 - z + \cdots = 2 - z + \cdots\)

 

[fresh_42]

The question looks as if my post #2 was correct and you don't need a series for [imath] 1/z. [/imath] You need powers from [imath] -1 [/imath] to [imath] 2, [/imath] so I would use the first three terms of the series from [imath] e^x [/imath] and [imath] 1/(1+z^2). [/imath]

 

[logistic_guy]

The question looks as if my post #2 was correct and you don't need a series for [imath] 1/z. [/imath] You need powers from [imath] -1 [/imath] to [imath] 2, [/imath] so I would use the first three terms of the series from [imath] e^x [/imath] and [imath] 1/(1+z^2). [/imath]

\(\displaystyle \frac{e^z}{z(z^2 + 1)} = \frac{1}{z}(1 + z + \frac{z^2}{2} + \cdots)(1 - z^2 + z^4 + \cdots)\)

\(\displaystyle = (\frac{1}{z} + 1 + \frac{z}{2} + \cdots)(1 - z^2 + z^4 + \cdots)\)

\(\displaystyle = \frac{1}{z} - z + z^3 + 1 - z^2 + z^4 + \frac{z}{2} - \frac{z^3}{2} + \frac{z^5}{2} + \cdots\)

this don't look like the opwhere my mistake?

 

[fresh_42]

\(\displaystyle \frac{e^z}{z(z^2 + 1)} = \frac{1}{z}(1 + z + \frac{z^2}{2} + \cdots)(1 - z^2 + z^4 + \cdots)\)

\(\displaystyle = (\frac{1}{z} + 1 + \frac{z}{2} + \cdots)(1 - z^2 + z^4 + \cdots)\)

\(\displaystyle = \frac{1}{z} - z + z^3 + 1 - z^2 + z^4 + \frac{z}{2} - \frac{z^3}{2} + \frac{z^5}{2} + \cdots\)

this don't look like the opwhere my mistake?

You gave up too early (a), it is necessary to consider the first four terms of the product series (b), and you should pair the terms by powers of [imath] z [/imath] (c). If you do all these then you will get

[math]\begin{array}{lll} \dfrac{e^z}{1+z^2}&=\left(\displaystyle{\sum_{k=0}^\infty \dfrac{z^k}{k!}}\right) \cdot \left(\displaystyle{\sum_{j=0}^\infty (-1)^{j}z^{2j}}\right)\\[24pt] &=\displaystyle{\sum_{n=0}^\infty \left(\sum_{k+j=n}\dfrac{(-1)^j}{k!}z^{k+2j}\right)}\\[24pt] &=(0,0)+\left[(0,1)+(1,0)\right]+\left[(0,2)+(1,1)+(2,0)\right]+\left[(0,3)+(1,2)+(2,1)+(3,0)\right]+\ldots\\[20pt] &=1+\left[-z^2+z\right]+\left[z^4-z^3+\dfrac{1}{2}z^2\right]+ \left[-z^6+z^5-\dfrac{1}{2}z^{4}+\dfrac{1}{6}z^3 \right]+\ldots \\[20pt] &=1+z-\dfrac{1}{2}z^2 -\dfrac{5}{6}z^3\pm\ldots\\[24pt] \dfrac{e^z}{z(1+z^2)}&=\dfrac{1}{z}+1-\dfrac{1}{2}z-\dfrac{5}{6}z^2\pm\ldots \end{array}[/math]

 

[logistic_guy]

You gave up too early (a), it is necessary to consider the first four terms of the product series (b), and you should pair the terms by powers of [imath] z [/imath] (c). If you do all these then you will get

you tell me only to use three terms

you've a nice solution over there.
i'll try to use \(\displaystyle 4\) terms this time and i'll try to simplify

\(\displaystyle \frac{e^z}{z(z^2 + 1)} = \frac{1}{z}(1 + z + \frac{z^2}{2} + \frac{z^3}{6} + \cdots)(1 - z^2 + z^4 - z^6 + \cdots)\)

\(\displaystyle = (\frac{1}{z} + 1 + \frac{z}{2} + \frac{z^2}{6} + \cdots)(1 - z^2 + z^4 - z^6 + \cdots)\)

\(\displaystyle = \frac{1}{z} - z + z^3 - z^5 + 1 - z^2 + z^4 - z^6 + \frac{z}{2} - \frac{z^3}{2} + \frac{z^5}{2} - \frac{z^7}{2} + \frac{z^2}{6} - \frac{z^4}{6} + \frac{z^6}{6} - \frac{z^8}{6} + \cdots\)

i'll simplify up to degree \(\displaystyle 2\)

\(\displaystyle = \frac{1}{z} - z + 1 - z^2 + \frac{z}{2} + \frac{z^2}{6} + \cdots\)

\(\displaystyle = \frac{1}{z} + 1 - \frac{2z}{2} + \frac{z}{2} - \frac{6z^2}{6} + \frac{z^2}{6} + \cdots\)

\(\displaystyle = \frac{1}{z} + 1 - \frac{z}{2} - \frac{5z^2}{6} + \cdots\)

i get it correct this timewhat a lovely day

thank fresh_42 very much

 

[fresh_42]

you tell me only to use three terms

I used only three terms of the original series. But to get all combinations up to [imath] z^2 [/imath] we need [imath] z^3 [/imath] before multiplying with [imath] 1/z [/imath] and that means all index combinations from [imath] (0,0) [/imath] to [imath] (0,3),\ldots,(3,0) [/imath] so counting zero makes four. Sorry.

 

[logistic_guy]

I used only three terms of the original series. But to get all combinations up to [imath] z^2 [/imath] we need [imath] z^3 [/imath] before multiplying with [imath] 1/z [/imath] and that means all index combinations from [imath] (0,0) [/imath] to [imath] (0,3),\ldots,(3,0) [/imath] so counting zero makes four. Sorry.

you're super

appreciate it